Packet Substitution Solutions - Integration by Substitution...

Integration by Substitution Solutions
Note: in (1) through (6), we find the answer by setting du equal to the
derivative of the function, multiplied by dx .
1. u = 4 x + 1
du = 4 dx
2. u = x
4
+ 2
du = 4 x
3
dx
3. u = 3 x
2
+ 1
du = 6 xdx
4. u = ln( x )
du =
1
x
dx
5. u = 1 /x = x
- 1
du = - x
- 2
dx
6. u = e
x
+ 2
du = e
x
dx
7. u = 4 x + 1
du = 4 dx
du/ 4 = dx
R
(4 x + 1)
4
dx =
R
u
4
du
4
=
1
4
R
u
4
du =
1
4
·
u
5
5
+ C =
1
20
(4 x + 1)
5
+ C
8. u = - 2 x + 3
du = - 2 dx
- du/ 2 = dx
R
e
- 2 x +3
dx =
R
e
u
· -
du
2
= -
1
2
R
e
u
du = -
1
2
e
u
+ C = -
1
2
e
- 2 x +3
+ C
9. Note: in problems like this one, it may not be immediately obvious
what to set u equal to (e.g., here, you might set u = 5 x
3
or u = x
4
+2). You
can try trial and error, by setting u equal to something and trying to solve
1

2
the problem, and then starting over if you can’t cancel out all of the ‘ x ’s;
another tip is to look for a function whose derivative you see in the integral.
So if you set u = x
4
+ 2, the derivative is 4 x
3
, which is similar to the 5 x
3
we already see in the integral; this is a good indicator that u = x
4
+ 2 is the
right choice.